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    (Original post by ceeebs)
    Does anybody know what synoptic topics I should look at? I know Chem5 isn't usually toooo synoptic but just incase AQA are feeling extra nasty this year and want to give us a paper like Bio5 urgh.
    Was thinking of learning conditions and reagents for reactions eg. unit 2/4 but Im not sure whether it will just be a waste of time lol
    I hated bio5 too! stupid aqa! :mad:
    I'd say go over unit 2 a bit more than unit 4...there's not really much they could as us from unit 4 that's relevant to unit 5 if you get me...unit 2 has the redox reactions etc..
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    (Original post by SkyHunter)
    Hi, can somebody help me with this question please...
    Its question 2b in topic 14.3 of the nelson thornes book
    the equation given is: I2(aq)+2Br(aq)--->Br2(aq)+2I-
    the E value of Br2+2e- -->2Br- =+1.07
    the E value of I2+2e- --->2I- =+0.54

    I worked the emf out to be 0.53 but the answers at the back say its -0.53 :confused:

    help? please?
    It's probably wrong. I remember doing the questions for this bit a while ago and it seems as though there are mistakes in it.
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    (Original post by Sparkly-Star)
    When doing deltaG = deltaH - TdeltaS

    When should you divide by 1000? I am always confused about that bit.
    you divide by 1000 for the delta s value as entropy is measured in joules/kelvin/mole where as enthalpy; deltaH is measured in kJ
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    (Original post by Sparkly-Star)
    It's probably wrong. I remember doing the questions for this bit a while ago and it seems as though there are mistakes in it.
    I thought so too...there's too many mistakes in the nelson thornes books :mad:
    thanks though x
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    (Original post by SkyHunter)
    I thought so too...there's too many mistakes in the nelson thornes books :mad:
    thanks though x
    This is my least favourite chapter so I find it annoying that the NT book doesn't explain this chapter clearly. Especially 14.4.
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    (Original post by Sparkly-Star)
    This is my least favourite chapter so I find it annoying that the NT book doesn't explain this chapter clearly. Especially 14.4.
    I think everyone find the redox equilibria bit difficult and yeh... it's a bit useless at times...i'm using the CGP book for that bit, its not much better tbh (N)
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    Do people think that we should revise Mass spectroscopy/Infrared spectroscopy, as it is in the AS textbook...

    Also, I have a feeling that we may get some chemical tests -
    e.g how to tell the difference between BaCl2 and MgCl2 (Add H2SO4), with equations and observations to tell them apart. So I'm going to revise group 2 reactions, the redox equations of AS and basic calculations like q=mcdeltaT and PV = nRT. Gah, so much on top of all the transition metal stuff!
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    Do all the transition metal reactions with sodium carbonate result in 2 being formed, e.g 2 crco3, 3 h2o and 3 co3? Or is it just aluminum?
    Also, will we have to know the structure of en or EDTA?!
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    Please can someone explain visible spectrophotometry to me?
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    could somebody help on the titration question on the jan 2010 paper plz?
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    What do we actually need to know about electrochemical cells?
    And can anyone explain the effects of temp and conc on emf of cells? It says in the book they have an effect but doesn't explain how and what. Stupid Nelson Thorne -____-
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    (Original post by SkyHunter)
    Hi, can somebody help me with this question please...
    Its question 2b in topic 14.3 of the nelson thornes book
    the equation given is: I2(aq)+2Br(aq)--->Br2(aq)+2I-
    the E value of Br2+2e- -->2Br- =+1.07
    the E value of I2+2e- --->2I- =+0.54

    I worked the emf out to be 0.53 but the answers at the back say its -0.53 :confused:

    help? please?
    Since Br2+2e- -->2Br- is reversed, then the Br2/Br- electrode is negative and is on the left in cell representation. while I2 / I- electrode is positive and on the right in cell representation. emf = Eright - Eleft.

    If you need further explanation, quote me.
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    (Original post by Crowned Temmy)
    Since Br2+2e- -->2Br- is reversed, then the Br2/Br- electrode is negative and is on the left in cell representation. while I2 / I- electrode is positive and on the right in cell representation. emf = Eright - Eleft.

    If you need further explanation, quote me.
    Wrong.

    The more positive one (Br) is the positive electrode, and on the right, where reduction occurs.
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    (Original post by Purple_Plume)
    What do we actually need to know about electrochemical cells?
    And can anyone explain the effects of temp and conc on emf of cells? It says in the book they have an effect but doesn't explain how and what. Stupid Nelson Thorne -____-
    This explanation is from collin's student guide.Using daniel's cell as example, if the equation is

    Zn + Cu2+ ---------> Cu + Zn2+

    Increasing concentration of Zn or Cu2+ shifts equilibrium to the right (according to le chatelier's principle) in order to reduce the effect of the constraint hence the forward reaction is favoured. Anytime the forward reaction is favoured, emf increases.

    Increasing concentration of Cu or Zn2+ (which are the products) shifts equilibrium to the left in order to reduce the effect of the constraint hence the backward reaction is favoured. when the backward reaction is favoured, emf decreases.

    Effects of increasing temp: Any reaction with positive emf is exothermic. Increasing temp will cause equilibrium to shift to the left in the direction that cools the system. This favours backward reaction and emf decreases.

    The same info applies to decreasing the conc and temp but the opposite happens in each case: the main thing is: IF FORWARD REACTION IS FAVOURED, emf INCREASES. IF BACKWARD REACTION IS FAVOURED, emf DECREASES.

    I will be glad to provide you with further explanation if needed.
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    (Original post by Luke0011)
    Wrong.

    The more positive one (Br) is the positive electrode, and on the right, where reduction occurs.
    If you check the original equation, I2 is reduced to I- so should be on the right. while Br- is oxidised to Br2 so should be on the left.

    It is the equation that matters. However, in the real sense, the reaction is not feasible ( remember I2 can not displace Br- from its compound ... AS halogens).
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    (Original post by Crowned Temmy)
    If you check the original equation, I2 is reduced to I- so should be on the right. while Br- is oxidised to Br2 so should be on the left.

    It is the equation that matters. However, in the real sense, the reaction is not feasible ( remember I2 can not displace Br- from its compound ... AS halogens).
    Ah, apologies, i did not see the equation!
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    (Original post by Luke0011)
    Ah, apologies, i did not see the equation!
    No problem man.
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    (Original post by Crowned Temmy)
    This explanation is from collin's student guide.Using daniel's cell as example, if the equation is

    Zn + Cu2+ ---------> Cu + Zn2+

    Increasing concentration of Zn or Cu2+ shifts equilibrium to the right (according to le chatelier's principle) in order to reduce the effect of the constraint hence the forward reaction is favoured. Anytime the forward reaction is favoured, emf increases.

    Increasing concentration of Cu or Zn2+ (which are the products) shifts equilibrium to the left in order to reduce the effect of the constraint hence the backward reaction is favoured. when the backward reaction is favoured, emf decreases.

    Effects of increasing temp: Any reaction with positive emf is exothermic. Increasing temp will cause equilibrium to shift to the left in the direction that cools the system. This favours backward reaction and emf decreases.

    The same info applies to decreasing the conc and temp but the opposite happens in each case: the main thing is: IF FORWARD REACTION IS FAVOURED, emf INCREASES. IF BACKWARD REACTION IS FAVOURED, emf DECREASES.

    I will be glad to provide you with further explanation if needed.
    Ah, that makes sence. Thank you very much My book is utter rubbish I think when it comes to actually having the right stuff in =/
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    (Original post by Crowned Temmy)
    If you check the original equation, I2 is reduced to I- so should be on the right. while Br- is oxidised to Br2 so should be on the left.

    It is the equation that matters. However, in the real sense, the reaction is not feasible ( remember I2 can not displace Br- from its compound ... AS halogens).
    i'm confused!
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    Guys what specific things do we need to know from unit 1, 2 and 4 for this exam??
    Please quote me
    Will rep for helpful answers
 
 
 
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